kuangbin专题四 ： 最短路 I 题 Arbitrage

Posted 2020-10-13 0一叶0知秋0

 

kuangbin专题四 ： 最短路 I 题  Arbitrage

POJ 2240

Arbitrage is the use of discrepancies in currency exchange rates to transform one unit of a currency into more than one unit of the same currency. For example, suppose that 1 US Dollar buys 0.5 British pound, 1 British pound buys 10.0 French francs, and 1 French franc buys 0.21 US dollar. Then, by converting currencies, a clever trader can start with 1 US dollar and buy 0.5 * 10.0 * 0.21 = 1.05 US dollars, making a profit of 5 percent.

Your job is to write a program that takes a list of currency exchange rates as input and then determines whether arbitrage is possible or not.

Input

The input will contain one or more test cases. Om the first line of each test case there is an integer n (1<=n<=30), representing the number of different currencies. The next n lines each contain the name of one currency. Within a name no spaces will appear. The next line contains one integer m, representing the length of the table to follow. The last m lines each contain the name ci of a source currency, a real number rij which represents the exchange rate from ci to cj and a name cj of the destination currency. Exchanges which do not appear in the table are impossible.
Test cases are separated from each other by a blank line. Input is terminated by a value of zero (0) for n.

Output

For each test case, print one line telling whether arbitrage is possible or not in the format "Case case: Yes" respectively "Case case: No".

Sample Input

3
USDollar
BritishPound
FrenchFranc
3
USDollar 0.5 BritishPound
BritishPound 10.0 FrenchFranc
FrenchFranc 0.21 USDollar

3
USDollar
BritishPound
FrenchFranc
6
USDollar 0.5 BritishPound
USDollar 4.9 FrenchFranc
BritishPound 10.0 FrenchFranc
BritishPound 1.99 USDollar
FrenchFranc 0.09 BritishPound
FrenchFranc 0.19 USDollar

0

Sample Output

Case 1: Yes
Case 2: No

题意：先给 货币的种类 数量n 种， 然后给出 n 种货币的名字 ， 再给出 转换方式的数量 m ， 之后 m 种 具体的 货币转换关系 。 
　　　某一种转换关系 a b c ， 表示 货币 a 乘以 比例 b 为可以 换到的 货币 c 的数量 。
　　　问是否存在正环。存在输出 Yes 否则输出 No

思路： 数组存放 n 种货币名称 ， 数组下标 表示 给货币 编号 。
　　　　然后给其中一种货币的本金置为 1 ，其他为 0 ， 然后用 bellman_ford 算法判断正环。

1.bellman_ford()

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std ; 
#define maxn 1000
char str[maxn][maxn] ; 
double dis[maxn] ; 
struct node {
    int a , 
        b ; 
    double rate ; 
};

node edge[maxn] ; 
int n , m ; 
// 获得 货币 s 在 数组中的编号 
int getpos(char *s){
    for(int i=1 ; i<=n ; i++){
        if(strcmp(str[i] , s  ) == 0 ){
            return i ; 
        }
    }
}
// 输入 && 建边 
void init(){
    
    char a[maxn] , c[maxn] ; 
    double b ; 
    
    for(int i=1 ; i<=n ; i++){
        scanf(" %s" , str[i]) ; 
    }
    scanf("%d" , &m) ; 
    for(int i=1 ; i<=m ; i++){
        scanf("%s %lf %s" , a , &b , c ) ; 
        edge[i].a = getpos(a) ; 
        edge[i].b = getpos(c) ;
        edge[i].rate = b ; 
    }
}
// 判断正环 
bool bellman_ford(){
    for(int i=1 ; i<=n ; i++){
        dis[i] = 0 ; 
    }
    dis[1] =1 ; 
    
    for(int i=1 ; i<=n ; i++){
        for(int j=1 ; j<=m ; j++){
            if(dis[edge[j].a] * edge[j].rate > dis[edge[j].b]){
                dis[edge[j].b] = dis[edge[j].a] * edge[j].rate ; 
                
            }
        }
    }
    
    for(int i=1 ; i<= m ;i++){
        if(dis[edge[i].a] * edge[i].rate > dis[edge[i].b]){
            return true ; 
        }
    }
    return false ; 
}

int main(){
    int times = 0 ; 
    while(~scanf("%d" , &n) && n ){
        
        init() ; 
        printf("Case %d: " , ++ times) ; 
        if(bellman_ford()){
            printf("Yes\n") ; 
        } else {
            printf("No\n") ; 
        }
    }
    
    return 0 ; 
} 
 

 2.floyd()

#include <iostream>  
#include <string.h>  
using namespace std;  
#define MAXC 100  
#define MAXV 50  
  
double map[MAXV][MAXV];  
int n,m;  
  
void Input(){  
    char s[MAXV][MAXC],a[MAXC],b[MAXC];  
    int i,k,j;  
    double c;  
    for(i=0;i<=n;i++)  
        for(j=0;j<=n;j++)  
            if(i==j)  
                map[i][j]=1;  
            else  
                map[i][i]=0;  
    for(i=1;i<=n;i++) scanf("%s",s[i]);  
    scanf("%d\n",&m);  
    for(i=1;i<=m;i++){  
        scanf("%s %lf %s",a,&c,b);  
        for(j=1;j<=n;j++)  
            if(!strcmp(s[j],a)) break;  
        for(k=1;k<=n;k++)  
            if(!strcmp(s[k],b)) break;  
        map[j][k]=c;  
    }  
}  
  
void floyd(){  
    int i,j,k;  
    for(k=1;k<=n;k++)  
        for(i=1;i<=n;i++)  
            for(j=1;j<=n;j++)  
                if(map[i][k]*map[k][j]>map[i][j])  
                    map[i][j]=map[i][k]*map[k][j];  
}  
  
int main(){  
    int cas=1,i;  
    while(scanf("%d\n",&n) && n){  
        Input();  
        floyd();  
        printf("Case %d: ",cas++);  
        for(i=1;i<=n;i++)  
            if(map[i][i]>1) break;  
        if(i>n) printf("No\n");  
        else printf("Yes\n");  
    }  
    return 0;  
}